Classically compute ahead time if Lie Algebra is either polynomial finite or geometrically closed?

Given a set of $$N$$ operators $$\mathcal{O}$$ with a known set of Lie Algebra group multiplication rules $$\mathcal{G}$$ that can be programmed into a classical computer, is there a classical poly($$N$$) calculation that can be done to check if $$\mathcal{O}$$'s Lie Algebra is either polynomial finite (1) or geometrically closed (2) under $$\mathcal{G}$$ without computing all $$N!$$ group multiplications in a brute force manner?

(1) Polynomial finite closed means that the number of extra operators that appear when performing all $$N!$$ possible group multiplications that need to be stored in classical memory grows poly($$N$$). This means that there is a lot of common factors when performing all possible group multiplications.

(2) Geometrically closed means that the number of extra operators that must be stored in memory grows exp($$N$$), but the complex scalar weights that are multiplied into those operators exponentially shrink in value with each increasing number of group multiplications. Note: In order for $$\mathcal{O}$$ to have associated complex scalar weights, their operators need to have a defined method of normalizing them so one can compare their weights. I chose the word geometric here, because of its similarity to the convergence of a geometric series, however their might be a better word.

Definitions (1) and (2) are important for classical computing of quantum many-body problems. If given a task to solve a quantum mechanical problem in an exponentially large Hilbert space, it would be nice to know if one can compute either (1) its exactly solvable in polynomial time, or (2) the quantity to compute, such as an expectation value of an operator, can be computed up to $$M$$ digits of precision and all other higher order corrections from many group multiplications can be truncated as an approximation.

An example of geometrically closed:

Let $$n$$ be a positive integer, $$\mathcal{H}$$ be a discrete $$2^n$$ by $$2^n$$ Hilbert space, and in $$\mathcal{H}$$ their exists $$m=2n+1$$ Dirac matrices $$\gamma^{k}$$ that span $$\mathcal{H}$$ under matrix multiplication and addition that satisfy the properties $$\gamma^{j}\gamma^{k}+\gamma^{k}\gamma^{j}=2\delta_{j,k} I$$, $$\delta_{k,j}$$ is the Kronecker delta, and $$I$$ is the $$2^n$$ by $$2^n$$ identity matrix, and all operators in $$\mathcal{H}$$ have their complex scalar coefficients compared under the Hilbert-Schmidt distance $$|A,B|_{\text{HS}} = \sqrt{ \dfrac{\text{Tr}((A-B)(A-B)^{\dagger})}{\text{Tr}(I)} } \ ,$$ then a set of operators $$\mathcal{O}^k =(\delta_{k,0}) I+ \dfrac{(1-\delta_{k,0})}{4} \gamma^{k} \ ,$$ where $$k=0,1,\cdots,(m-2),(m-1)$$, "could be" geometrically closed under the matrix commutator $$[A,B]=AB-BA$$, because any evalation of $$r$$ number of commutators on the set $$\mathcal{O}^k$$ excluding $$\mathcal{O}^0$$ of the form $$\Bigg[\mathcal{O}^{k_0}, \bigg[\mathcal{O}^{k_1}, \Big[\mathcal{O}^{k_2}, \big[\mathcal{O}^{k_3}, \cdots \big] \Big] \bigg] \Bigg] =M_r$$ has the relative distance measure $$|\mathcal{O}^0,\mathcal{O}^0+M_r|_{\text{HS}} \approx \dfrac{1}{2^r}$$ which exponentially shrinks in scalar value with increasing number of commutators $$r$$.

There "could be" a problem in this example when there is significant overlap of commutators under matrix addition which could sum up to values that do not exponentially shrink under $$| \ |_{\text{HS}}$$.

• There are standard definition in this context: for a finitely generated algebra (often assumed Lie or associative), generated by a finite-dimensional subspace $V$, the growth function is $n\mapsto b(n)=\dim(V+V^2+\dots +V^n)$. Here $V^k$ is the subspace generated by all possible products of $k$ elements of $V$ (in a non-associative context one takes all possible parentheses). The asymptotics of $b$ does not depend on the choice of $V$. Does "polynomially finite" mean that $b$ is bounded above by a polynomial? Is "geometrically closed" related to the growth of $b$?
– YCor
Sep 23 '19 at 14:31
• @YCor Yes. For polynomial finite, $b$ is bounded above by a polynomial growth in n. For geometrically closed, b(n) is exponential, but the complex scalar coefficients associated with each operator exponentially shrink in k from $V^k$. It sounds like your implying a reference that proves this, if so please send it. Sep 23 '19 at 14:56
• No I have no such claim, I'm just trying to understand the definition (I don't understand that of "geometrically closed").
– YCor
Sep 23 '19 at 14:58
• @YCor I wrote an example in my OP to provide clarity to the definition of geometrically closed. Please let me know if there is any confusion. Sep 23 '19 at 16:00
• From it I realize that the setting itself is not that clear to me. It seems that you're considering the Lie $\mathbf{Z}$-algebra generated by a finite subset of a matrix algebra, and the definition of "geometrically closed" is related to the norms of the given matrices.
– YCor
Sep 24 '19 at 6:27